Question

Fine an equation in y-intercept form for the tangent line to the curve x^2-y^2=1 at the point (sqrt(2),1)

Answer 1

Can someone show me how to do this step by step??

Answer 2

So we're trying to construct a line:\[\Large\rm y=\color{orangered}{m}x+b\]

Answer 3

This line is tangent to our curve, so the `slope` of our line will be given by the derivative of our function evaluated at the particular point of tangency (is that a word? I dunno).

\[\Large\rm y=\color{orangered}{y'\left(\sqrt2\right)}x+b\]Ooo that's going to get confusing.
Let's call our tangent line y, Y instead.\[\Large\rm Y=\color{orangered}{y'\left(\sqrt2\right)}x+b\]

Answer 4

So to get the slope of our tangent line we need to take the derivative of our function, then evaluate it at x=sqrt2, y=1,
that will give us our slope.

Do you understand how to differentiate this function?\[\Large\rm x^2-y^2=1\]

Answer 5

I believe so. is dy/dx=-y/x ??

Answer 6

\[\Large\rm 2x-2yy'=0\qquad\implies\qquad -yy'=-x\]Hmm maybe you missed a negative in there? :o

Answer 7

Oh, whoops! So is it y/x?

Answer 8

Hmm I think our division might be backwards also :3
We would end up dividing both sides by -y, yes?
so y ends up on the bottom?

Answer 9

ohhhh yeah!

Answer 10

dy/dx= x/y

Answer 11

Ok great :)

And our derivative is a function of `both` x and y,\[\Large\rm y'(x,y)=\frac{x}{y}\]

Answer 12

Now we want to evaluate this function at the given point,
the result will be the `slope` of our tangent line.

Answer 13

so does x/y = (sqrt(2))/1)

Answer 14

= sqrt(2)

Answer 15

Mmm ok good.

I made a little boo boo earlier,
I was assuming our derivative would be a function of x, so I wrote this:\[\Large\rm Y=\color{orangered}{y'\left(\sqrt2\right)}x+b\]But it's really this:\[\Large\rm Y=\color{orangered}{y'\left(\sqrt2,1\right)}x+b\]And you determined that the derivative at that point is sqrt(2),\[\Large\rm Y=\color{orangered}{\sqrt2~}x+b\]

Answer 16

oh okay! no worries :)

Answer 17

So now to find our y-intercept, \(\Large\rm b\), realize that both our curve and tangent line share the point (sqrt2,1), so we can plug that into our tangent line to solve for b.

Answer 18

\[\Large\rm (\sqrt2,1) \qquad\to\qquad \Large\rm Y=\sqrt2~x+b\]

Answer 19

Plug it in :d what do you get for b?

Answer 20

B= -1 ?

Answer 21

Ok great! ^^

Answer 22

So that gives us all the information we needed.

Answer 23

\[\Large\rm Y=\sqrt2~x-1\]

Answer 24

yay team \c:/

Answer 25

THANK YOU THANK YOU THANK YOU!!! :)))

Answer 26

Here is a picture of the graph, in case it helps to visualize it:
https://www.desmos.com/calculator/qhtxwhguw1

Answer 27

Gotcha! Cool cool! You're the best!